A consumer advocacy group feels that Walmart provides a less safe shopping envir

Question

A consumer advocacy group feels that Walmart provides a less safe shopping environment than Target. To try to prove their point, they randomly selected 20 Walmart stores and found the annual number of crime reports filed for these stores. Each of these Walmart stores was then paired with a Target store within a 10 mile radius, and the annual crime reports for the associated Target stores were also recorded. The data is shown below. Test to to see if the mean number of crime incidents at Walmart stores is significantly larger than the mean number of crime incidents at Target stores. Please note the data shown below is not real, but if you would like to see the real example, check out walmartcrimereport.com. Unfortunately, the real data was collected in a way that was not fair to Walmart. Assume normality.

a) Let W represent the mean incidents associated with Walmart, and let T represent the mean incidents associated with Target. What are the proper hypotheses?

H0: W = T versus Ha: W T

H0: W = T versus Ha: W = T    

H0: W = T versus Ha: W > T

H0: W = T versus Ha: W <

b) What is the test statistic? Give your answer to four decimal

c) Using a 0.05 level of significance, what is the critical point for this test? Give your answer to four decimal places.  
d) What is the appropriate conclusion?

Conclude that W is greater than T because the test statistic is larger than the critical point.

Fail to reject the claim that W is equal to T because the test statistic is larger than the critical point.     

Reject the claim that W is greater than T because the test statistic is larger than the critical point.

Fail to reject the claim that W is equal to T because the test statistic is less than the critical point.

Explanation / Answer

Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 > u2
Test Statistic
X(Mean)=815.55
Standard Deviation(s.d1)=141.3344 ; Number(n1)=20
Y(Mean)=781.55
Standard Deviation(s.d2)=146.3688; Number(n2)=20
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =815.55-781.55/Sqrt((19975.41262/20)+(21423.82561/20))
to =0.747
| to | =0.747
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 19 d.f is 1.729
We got |to| = 0.7473 & | t | = 1.729
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Right Tail -Ha : ( P > 0.7473 ) = 0.23201
Hence Value of P0.05 < 0.23201,Here We Do not Reject Ho

[ANSWERS]
1. H0: W = T versus Ha: W > T
2. to =0.747
3. Critical Value, 1.729
4. Fail to reject the claim that W is equal to T because the test statistic
is less than the critical point.

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